# If sin 2 </msup> &#x2061;<!-- ⁡ --> x + sin 2 </msup>

If ${\mathrm{sin}}^{2}x$ + ${\mathrm{sin}}^{2}2x$ + ${\mathrm{sin}}^{2}3x$ = 1, what does ${\mathrm{cos}}^{2}x$ + ${\mathrm{cos}}^{2}2x$ + ${\mathrm{cos}}^{2}3x$ equal?
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Darrell Valencia
You know that
${\text{sin}}^{2}x+{\text{sin}}^{2}2x+{\text{sin}}^{2}3x=1\phantom{\rule{2em}{0ex}}\left(\text{Eq.}1\right)$
Notice that
$\begin{array}{r}{\text{sin}}^{2}x+{\text{cos}}^{2}x=1⇒{\text{sin}}^{2}x=1-{\text{cos}}^{2}x\end{array}$
Substituting the identity in Equation (1):
$\begin{array}{rl}\underset{1-{\text{cos}}^{2}x}{\underset{⏟}{{\text{sin}}^{2}x}}& +\underset{1-{\text{cos}}^{2}2x}{\underset{⏟}{{\text{sin}}^{2}2x}}+\underset{1-{\text{cos}}^{2}3x}{\underset{⏟}{{\text{sin}}^{2}3x}}=1\\ \\ & ⇒1-{\text{cos}}^{2}x+1-{\text{cos}}^{2}2x+1-{\text{cos}}^{2}3x=1\\ & ⇒3-\left({\text{cos}}^{2}x+{\text{cos}}^{2}2x+{\text{cos}}^{2}3x\right)=1\\ & ⇒2={\text{cos}}^{2}x+{\text{cos}}^{2}2x+{\text{cos}}^{2}3x\end{array}$
Therefore:
${\text{cos}}^{2}x+{\text{cos}}^{2}2x+{\text{cos}}^{2}3x=2$
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Yesenia Obrien
${\mathrm{sin}}^{2}x+{\mathrm{sin}}^{2}2x+{\mathrm{sin}}^{2}3x=1\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\left(1-co{s}^{2}x\right)+\left(1-{\mathrm{cos}}^{2}2x\right)+\left(1-co{s}^{2}3x\right)=1$
Rearranging terms we get ${\mathrm{cos}}^{2}x+{\mathrm{cos}}^{2}2x+{\mathrm{cos}}^{2}3x=2$